Đáp án:

Giải thích các bước giải:

\(\eqalign{ & 1)\,\,y = \left( {2m - 5} \right)x + 3 \cr & a)\,\,Ham\,\,so\,la\,ham\,\,bac\,\,nhat \Leftrightarrow 2m - 5 \ne 0 \Leftrightarrow m \ne {5 \over 2} \cr & b)\,\,Ham\,\,so\,\,DB \Leftrightarrow 2m - 5 > 0 \Leftrightarrow m > {5 \over 2} \cr & \,\,\,\,\,\,Ham\,\,so\,\,NB \Leftrightarrow 2m - 5 < 0 \Leftrightarrow m < {5 \over 2} \cr} \) \(\eqalign{ & 2)\,\,Xet\,\,\Delta AHC: \cr & AH = AC.\sin {40^0} \approx 6,23\,\,\left( {cm} \right) \cr & Xet\,\,\Delta AH: \cr & BH = AH.\cot {50^0} \approx 5,4\,\,\left( {cm} \right) \cr} \) \(\eqalign{ & 3)\,\,a)\,\,Ke\,\,AH \bot BC \cr & Xet\,\,{\Delta _v}AHC:\,\,AH = AC.\sin {50^0} \approx 26,81\,\,\left( {cm} \right) \cr & Xet\,{\Delta _v}AHB:\,\,\, \cr & \sin {60^0} = {{AH} \over {AB}} \Rightarrow AB = {{AH} \over {\sin {{60}^0}}} \approx 30,96\,\,\left( {cm} \right) \cr & Ap\,\,dung\,\,DL\,\,pytago\,\,trong\,\,{\Delta _v}ABC: \cr & BC = \sqrt {A{B^2} + A{C^2}} \approx 46,72\,\,\left( {cm} \right) \cr & b)\,\,{S_{\Delta ABC}} = {1 \over 2}AH.BC \approx 626,39\,\,\left( {c{m^2}} \right) \cr} \)

Đáp án:

1

)

y

=

(

2

m

−

5

)

x

+

3

a

)

H

a

m

s

o

l

a

h

a

m

b

a

c

n

h

a

t

⇔

2

m

−

5

≠

0

⇔

m

≠

5

2

b

)

H

a

m

s

o

D

B

⇔

2

m

−

5

>

0

⇔

m

>

5

2

H

a

m

s

o

N

B

⇔

2

m

−

5

<

0

⇔

m

<

5

2

2

)

X

e

t

Δ

A

H

C

:

A

H

=

A

C

.

sin

40

0

≈

6

,

23

(

c

m

)

X

e

t

Δ

A

H

:

B

H

=

A

H

.

cot

50

0

≈

5

,

4

(

c

m

)

3

)

a

)

K

e

A

H

⊥

B

C

X

e

t

Δ

v

A

H

C

:

A

H

=

A

C

.

sin

50

0

≈

26

,

81

(

c

m

)

X

e

t

Δ

v

A

H

B

:

sin

60

0

=

A

H

A

B

⇒

A

B

=

A

H

sin

60

0

≈

30

,

96

(

c

m

)

A

p

d

u

n

g

D

L

p

y

t

a

g

o

t

r

o

n

g

Δ

v

A

B

C

:

B

C

=

√

A

B

2

+

A

C

2

≈

46

,

72

(

c

m

)

b

)

S

Δ

A

B

C

=

1

2

A

H

.

B

C

≈

626

,

39

(

c

m

2

)

Giải thích các bước giải: